When a Product Is a Square or Prime Times a Square

Factor:

Let d = gcd(x - y, x + y). Since d divides both x - y and x + y,

Since gcd(x, y) = 1, we get d = 1 or 2.

Also,

so x - y is odd. Therefore,

We know

So x - y and x + y should both be perfect squares.

The possible pairs are

Since

the minimum occurs for (a, b) = (5, 9):

For p = 2, the expression is not an integer. For p = 3,

so p = 3 works. Now assume p ≥ 5.

Let

so d divides 2. Since both numbers are odd, d = 1.

Therefore there are two cases.

Case 1:

Since gcd(y - 1, y + 1) divides 2, and both factors are powers of 2, write

Their gcd is 2, so 2^a = 2. Hence y = 3.

Case 2:

Since p ≥ 5,

But a square is congruent to 0 or 1 modulo 4, so w² + 1 is congruent to 1 or 2 modulo 4, a contradiction.

Therefore,